The circuit can be analysed 'by hand', rather than using SPICE, to better understand its operation.

The 3.3V reference (V_{ref}) is applied to the + input of amplifier A. Assuming a perfect OPAMP, feedback through the PRT (R_{T}) makes the voltage at its - input also equal to V_{ref}. This appears across the 3.3k resistor (R_{ref}), causing a current V_{ref}/R_{ref} of nominally 1mA (I_{ref}) to flow through it.

By Kirchoff's Current Law, I_{ref} must be equal to the sum of the current through the PRT (I_{T}) and the 'compensation' current through the 47k resistor (which I label R_{3}). Denote the output of amplifier A as V_{T}. The 4.7k/100k divider network multiplies this by a factor *k* (*k* < 1) and applies it to the + input of amplifier B. Feedback round this amplifier holds its - input at the same voltage, namely *k*V_{T}.

Now the current through the PRT, I_{T} is given by:

I_{T} = (V_{T} - V_{ref})/R_{T}

and the compensation current I_{comp} through R_{3} is given by:

I_{comp} = (*k*V_{T} - V_{ref})/R_{3}

Equating the sum of the two currents to I_{ref} gives:

V_{ref}/R_{ref} = (V_{T} - V_{ref})/R_{T} + (*k*V_{T} - V_{ref})/R_{3}

Solving for V_{T} gives (after multiplying numerator & denominator by R_{T}, for clarity):

V_{T} = V_{ref}(R_{T}(1/R_{3} + 1/R_{ref}) + 1)/(R_{T}*k*/R_{3} + 1)

Note that R_{T} appears in both numerator and denominator, but its multiplying factor in the denominator is much larger.

This voltage is attenuated by the factor *k* and then amplified in the output stage, a non-inverting amplifier referenced to V_{ref}, so the final output voltage is:

V_{out} = (*k*V_{T} - V_{ref})(R_{3} + R_{4})/R_{3} + V_{ref}

where R_{4} is the 100k feedback resistor.

Try a few values of R_{T} near 100 ohms and see how the output voltage varies:

100.00 ohms (0˚C) gives 3.13574V

98.04 ohms (-5˚C) gives 3.12989V

101.95 ohms (+5˚C) gives 3.14156V

1.164 mV/˚C for one interval and 1.170 for the other, which is acceptable, if a little insensitive. The OPAMP offset voltages may cause problems, though. What happens to the output voltages in the rest of the system?